Question

$\huge{\tt{Ello!!!}}$
❄️A particle executes SHM on a straight line path. The amplitude of oscillation is 2 cm. When the displacement of the particle from the mean postion is 1 cm,the magnitude of its acceleration is equal to that of velocity. Find the time period of SHM, also the max. velocity and max. acceleration of SHM.
#all the best ♡​

Answer 1

Answer:

Given:

Amplitude = 2 cm

When displacement of particle from mean position is 1 cm, the acceleration and velocity are same.

To find:

• Time period
• Max Velocity
• Max acceleration

Calculation:

Let displacement be 1 cm,

Here I am considering magnitude of acceleration.

$\therefore \: acc. = velocity$

$= > { \omega}^{2} x = \omega \sqrt{ {A}^{2} - {x}^{2} }$

$= > \omega \: x = \sqrt{ {A}^{2} - {x}^{2} }$

Squaring on both sides:

$= > { (\omega \: x)}^{2} = {A}^{2} - {x}^{2}$

Putting x = 1 cm and A = 2 cm

$= > {( \omega \times 1)}^{2} = {2}^{2} - {1}^{2}$

$= > { \omega}^{2} = 3$

$= > \omega = \sqrt{3}$

Hence,

$time \: period = \dfrac{2\pi}{ \omega}$

$= > time \: period = \dfrac{2\pi}{ \sqrt{3} }$

Now,

$max \: acc. \: = - { \omega}^{2} A \\ = > max \: acc. = - 3 \times 2 \\ = > max \: acc. = - 6 \: cm \: {s}^{ - 2}$

Negative sign denotes opposite direction of acceleration with respect to displacement.

Now,

$max \: vel. = \omega \times A \\ = > max \: vel. = \sqrt{3} \times 2 \\ = > max \: vel. = 2 \sqrt{3} \: cm \: {s}^{ - 1}$

Answer 2

Answer:

T = 2 π / √ 3

v_max =  2 √ 3 m / sec

a_max = - 6 m / sec²

Explanation:

Given :

Amplitude A = 2 cm

Displacement of particle from mean position x = 1 cm

It is said acceleration is equal to velocity of particle.

ω ( √ A² - x² ) =  ω² x

ω x = ( √ A² - x² )

Putting value here :

ω  = ( √ 4 - 1 )

ω = √ 3

We know :

ω = 2 π / T

T = 2 π / √ 3 .

Now for :

v_max = A ω

v_max = 2 × √ 3

v_max =  2 √ 3 m / sec

For a_max = - A ω²

a_max = - 2 ( √ 3 )²

a_max = - 6 m / sec²

Hence we get answer.